Question

use the quotient rule to find the derivative of the following.
y = \frac{9x^{2}+2}{x^{2}+1}
y = \square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 9x^{2}+2$, $v=x^{2}+1$.

Step2: Find $u'$ and $v'$

Differentiate $u = 9x^{2}+2$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $u'=\frac{d}{dx}(9x^{2}+2)=18x$. Differentiate $v=x^{2}+1$ with respect to $x$, we get $v'=\frac{d}{dx}(x^{2}+1)=2x$.

Step3: Apply quotient - rule

Substitute $u$, $u'$, $v$, and $v'$ into the quotient - rule formula:
\[

$$\begin{align*} y'&=\frac{(18x)(x^{2}+1)-(9x^{2}+2)(2x)}{(x^{2}+1)^{2}}\\ &=\frac{18x^{3}+18x-(18x^{3}+4x)}{(x^{2}+1)^{2}}\\ &=\frac{18x^{3}+18x - 18x^{3}-4x}{(x^{2}+1)^{2}}\\ &=\frac{14x}{(x^{2}+1)^{2}} \end{align*}$$

\]

Answer:

$\frac{14x}{(x^{2}+1)^{2}}$